A box of 11 transistors has 4 defective ones.
A) If 2 transistors are drawn from the box together, what is the probability that both transistors are defective?
B) If 2 transistors are drawn from the box together, what is the probability that neither transistor is defective?
C) If 2 transistors are drawn from the box together, what is the probability that one transistor is defective?

Answers

Answer 1

Answer:

(a) 0.1325

(b) 0.4045

(c) 0.463

Step-by-step explanation:

Let X denote the number of defective transistors.

The proportion of defective transistors is, p = 4/11 = 0.364.

All the transistors are independent of the others.

The random variable X follows a binomial distribution.

(a)

Compute the probability that both transistors are defective, if 2 transistors are drawn from the box together as follows:

[tex]P(X=2)={2\choose 2}(0.364)^{2}(1-0.364)^{2-2}\\\\=1\times 0.132496\times 1\\\\=0.132496\\\\\approx 0.1325[/tex]

(b)

Compute the probability that neither transistors are defective, if 2 transistors are drawn from the box together as follows:

[tex]P(X=0)={2\choose 0}(0.364)^{0}(1-0.364)^{2-0}\\\\=1\times 1\times 0.404496\\\\=0.404496\\\\\approx 0.4045[/tex]

(c)

Compute the probability that one transistors are defective, if 2 transistors are drawn from the box together as follows:

[tex]P(X=1)={2\choose 1}(0.364)^{1}(1-0.364)^{2-1}\\\\=2\times 0.364\times 0.636\\\\=0.463008\\\\\approx 0.463[/tex]


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